Lecture XXVI. Morris Swartz Dept. of Physics and Astronomy Johns Hopkins University November 5, 2003

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1 Lecture XXVI Morris Swartz Dept. of Physics and Astronomy Johns Hopins University November 5, 2003

2 Lecture XXVI: Oscillations Oscillations are periodic motions. There are many examples of them in nature: vibrating strings (in musical instruments), drums, bells, etc vibrating pressure in gases, liquids, and solids (sound) vibrating electromagnetic waves - radio, light, x-rays, gamma-rays,... Most oscillating systems lose energy to their environments: damped oscillations. We can restore energy to such systems by externally driving them: driven oscillations. The rate at which an oscillation repeats is called its frequency f, f = Number of oscillations time In SI units, frequency is measured in hertz, 1 hertz = 1 Hz = 1 oscillation sec = 1 s 1

3 The time that it taes for one complete oscillation to occur is the period T T = 1 f and is measured in seconds. Periodic motion which can be described by the sine or cosine functions is called Simple Harmonic Motion (SHM). Eg, suppose that the displacement of a particle (or the com of an extended object) moves in 1-d with SHM, then we can write its displacement as a function of time as displacement at time t phase x(t) = A cos( t+ ) y A = t+ x(t) amplitude x angular frequency phase constant Note that this loos much lie the x position of an object rotating at constant angular speed ω where A is the length of the rotor which maes an angle θ = ωt + φ wrt the x-axis.

4 It is clear that φ is the angle of the rotor at time t = 0 (defines x(0)). If we set φ = 0, we can plot x(t) versus ωt, A -A /2 T 3 /2 2 t We see that A is the maximum amplitude of the oscillation and that a full period occurs when, ωt = 2π T = 2π ω the period is inversely related to ω. Using our definition of frequency, f = 1 T = ω 2π ω = 2πf the angular frequency is the product of the number of radians in one period and the frequency. It is clear that the system oscillates faster as ω is increased.

5 The effect of varying the parameters of x(t) is shown below, x i O x T A t We can as what is the velocity of the object moving with SHM? We are describing a 1-d system and we now how to define v, v i O v v max = ωa ω t v = dx dt = ωa sin(ωt + φ) = v max sin(ωt + φ) O a a max = ω 2 A t we see that the velocity is also oscillating with an amplitude v max = ωa but shifted by 1/4 period in phase.. We see that v = 0 where x is maximum and v is maximum where x = 0!

6 We can also calculate the acceleration of the object by taing another derivative, a = dv dt = d2 x dt 2 = ω 2 A cos(ωt + φ) = a max cos(ωt + φ) which we can also express as a(t) = ω 2 x(t). We see that the acceleration is also oscillating with an amplitude a max = ω 2 A but is a full 1/2 period out of phase with the displacement. We can as what inds of systems display SHM? The answer is: any system that has restoring force that is proportional to the displacement away from its equilibrium position will oscillate with Simple Harmonic Motion. As an example, consider a mass moving on a frictionless surface coupled to a spring, We now that the restoring force is F s = x.

7 Newton s 2 nd Law ma = F is therefore m d2 x dt 2 = x d2 x dt 2 + m x = 0 }{{} differential equation This equation (a 2 nd -order linear homogeneous differential equation) describes a system in simple harmonic motion. It is easy to see that x(t) = A cos(ωt + φ) is a solution of this equation, d 2 dt 2 [A cos(ωt + φ)] + m [A cos(ωt + φ)] = 0 Aω 2 cos(ωt + φ) + A cos(ωt + φ) = 0 m ( ω 2 + ) A cos(ωt + φ) = 0 m the left-hand side of the equation will be zero at all times if and only if, ω 2 + m = 0 ω = m We see that SHM is a solution of the equation as long as it has the form ( ) x(t) = A cos t m + φ

8 Our spring and mass will obey SHM with the angular frequency fixed to be ω = /m. This implies that the period and frequency are T = 2π ω = 2π m f = 1 T = 1 2π m Note that ω (and f) increase with and decrease with 1/ m. Stiffer springs oscillate faster and smaller masses oscillate faster. Note also that A and φ are not fixed by our equation (Newton s 2 nd Law): The frequency of an oscillator in SHM is independent of the amplitude of the oscillation! The amplitude and phase of the oscillation are determined by the initial conditions x(0) and v(0). For example, assume that we release the mass from x(0) with velocity v(0), x(0) = A cos(φ) v(0) = ωa sin(φ) Taing the ratio of these expressions, v(0) x(0) = ω tan(φ) tan(φ) = 1 v(0) ω x(0)

9 and taing the sum of the squares, x 2 (0) + A = ± ( ) v(0) 2 = A 2 (cos 2 φ + sin 2 φ) = A 2 ω x 2 (0) + v2 (0) ω 2 We see that A and φ are determined by the initial displacement and velocity. Two common special cases are: 1. Mass is released from rest with non-zero displacement (v(0) = 0): tan φ = 0 φ = 0, A = x(0) x(t) = x(0) cos (ωt) 2. Mass has zero initial displacement and non-zero initial velocity (x(0) = 0): tan φ = φ = π 2, A = v(0)/ω x(t) = v(0) ( ω cos ωt π ) = v(0) sin (ωt) 2 ω

10 We can easily analyze the inetic and potential energies of the mass-spring system in SHO. The potential energy is given by our well-nown expression, U(t) = 1 2 x2 (t) = 1 2 A2 cos 2 (ωt + φ) it oscillates from 0 to A 2 /2. The inetic energy is also given by a well-nown expression, K(t) = 1 2 mv2 (t) = 1 2 mω2 A 2 sin 2 (ωt + φ) and oscillates from 0 to ma 2 ω 2 /2. Summing the inetic and potential energies to calculate the mechanical energy, it is useful to note that mω 2 =, E(t) = K(t) + U(t) E(t) = 1 2 mω2 A 2 sin 2 (ωt + φ) A2 cos 2 (ωt + φ) = 1 2 A2 sin 2 (ωt + φ) A2 cos 2 (ωt + φ) = 1 2 A2

11 The mechanical energy is a constant in time (as it is required to be by the conservation of energy). This is shown graphically for K and U as functions of time and displacement. Note that E depends upon A which in turn depends upon the initial conditions. The energy of the system is what we put into it and cannot depend upon the oscillation frequency which is a fixed feature of the system.

12 L θ s mg sin T θ θ m g m mg cos θ Another system with an (almost) linear restoring force is the simple pendulum. A simple pendulum is a mass m hung from a massless rod or string of length L. When the pendulum is displaced from the vertical by an angle Θ, gravity provides a restoring torque τ (the sign indicates that τ always acts to mae Θ smaller), τ = LF = Lmg sin Θ Newton s 2 nd Law Iα = τ, becomes, I d2 Θ dt = Lmg sin Θ d2 Θ 2 dt 2 + Lmg I sin Θ = 0 This equation does not match our magic form for SHM! Note however, that for small displacements, sin Θ Θ. Maing this approximation and noting that I = ml 2, our equation becomes, d 2 Θ dt 2 + g L Θ = 0 this equation is now identical to those we ve already

13 considered. We can write-down the solution immediately, Θ(t) = A Θ cos(ωt + φ) ω = g L where the angular frequency depends only upon g and the length of the pendulum. For small angular displacements, the pendulum moves in SHM with a period of T = 2π ω = 2π L g Note that long pendula move more slowly than short ones and that a gravitational pendulum is a very convenient way to measure g. One must be very careful with notation when discussing the angular velocity of the pendulum. By definition it is, Ω(t) = dθ dt = ωa Θ sin(ωt + φ) = Ω max sin(ωt + φ) The angular velocity of the rotor Ω(t) and its amplitude Ω max = ωa Θ are NOT the frequency of the oscillation ω. The angular velocity oscillates bac and forth from Ω max to +Ω max which depend upon the initial conditions. The frequency of the oscillation ω does not depend upon initial conditions.

14 CPS Question 1 A bloc of mass M is hung from two identical springs of spring constant in parallel [see (b)]. M (a) M (b) How does its oscillation frequency compare with that for a bloc suspended from a single spring of the same spring constant [see (a)]? (A) ω b < ω a (B) ω b = ω a (C) ω b > ω a

15 The force produced by a single spring is F = x for a displacement of x from the equilibrium position. The frequency of the single spring oscillator is ω = /m. For the double spring oscillator, each spring produces a force F = x, x=0 F F F=x x M M (a) (b) so the net force is F net = 2x and the frequency of the two spring system is ω b = 2 m > m = ω a The correct answer is (C)!

16 CPS Question 2 A bloc of mass M is hung from two identical springs of spring constant which are connected end-to-end (in series) [see (b)]. M (a) How does its oscillation frequency compare with that for a bloc suspended from a single spring of the same spring constant [see (a)]? M (b) (A) ω b < ω a (B) ω b = ω a (C) ω b > ω a

17 F=x/2 x/2 M x/2 x=0 (b) The correct answer is (A)! x A total displacement of the end-to-end double spring of x displaces each spring by x/2. Therefore the tension force generated by each spring is F = x/2 and since only one spring is attached to the bloc, the total force on the bloc is also F = x/2. The frequency of the series spring system is thus ω b = 2m < m = ω a

18 Problem Two blocs (m = 1.0 g and M = 10 g) and a spring ( = 200 N/m) are arranged on a horizontal frictionless surface. The coefficient of static friction between the blocs is What amplitude of simple harmonic motion of the spring blocs system puts the smaller bloc on the verge of slipping over the larger bloc? Assuming that the two blocs are stuc together, they oscillate with an angular frequency ω = M + m = 200 N/m 11 g = 4.26 s 1 which means that the bloc pair accelerates with a time dependence a(t) = ω 2 A cos(ωt + φ)

19 and the pea acceleration is a max = ω 2 A where A is the amplitude of the oscillation. The maximum force that the lower bloc can exert on the upper bloc before sliding begins is F max = µ s N = µ s mg. The maximum acceleration that can be sustained is thus a max = F max m = µ sg Setting the maximum acceleration to the pea acceleration and solving for A, we find the maximum amplitude of the oscillation before sliding begins µ s g = ω 2 A A = µ sg ω 2 = 0.22 m